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Manhattan Ratios Problem

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Manhattan Ratios Problem

by kakz » Mon Oct 31, 2011 10:44 am
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?
(A)15%
(B)25%
(C)35%
(D)45%
(E)55%
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by ripulgupta » Mon Oct 31, 2011 10:59 am
Ratio of A:B remains the same so we retain 4:7:C:D.
Ration of A:C is made 4 times. So 4:8 becomes 4:2 Thus 4:7:2:D.
For B to be 20% of the mixture 7/(4+7+2+D) = .2. Thus D is 22

So A:D changes from 4:12 to 4:22 So the 4/12 needs to be multiplied by 12/22 which is just more than 50% so the ratio decreased by just below 50% answer should be D.
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by GMATGuruNY » Mon Oct 31, 2011 11:14 am
kakz wrote:The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?
(A)15%
(B)25%
(C)35%
(D)45%
(E)55%
Since A:C = 4:8, A:C quadrupled = 12:8 = 4:2.
A:B held constant = 4:7.
Thus, new A:B:C = 4:7:2.

Since B = 1/5 of the new mixture:
7 = (1/5)x.
x = 35.

Since x=35 and A+B+C = 4+7+2 = 13:
D = 35-13 = 22.

Old A : D = 4:12 = 1:3 = 11/33.
New A : D = 4:22 = 2:11 = 6/33.

Since old A : D and new A : D have the same denominator, we can determine the percent decrease using only the numerators:
(11-6)/11 * 100 = 500/11 ≈ 45.

The correct answer is D.
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